calculus - prove that a function is continuous if the denominator is a continuous function

Given that $f(x)$ is a continuous function and its value is positive at point $x_0$ I need to prove that:
$$f(x) + \frac{1}{f(x)}$$
is continuous at $x_0$. I think this is really easy but not 100% sure. My thinking is:
1) because $f(x)$ is continuous at $x_0$ then $lim_{x \to x_0} f(x)$ exists (definition of a continuous function at a given point).

2) Let's see if $lim_{x \to x_0} f(x) + \frac{1}{f(x)}$ exists: $lim_{x \to x_0} f(x) + \frac{1}{f(x)} = lim_{x \to x_0} \frac{(f(x))^2 +1}{f(x)}$, provided that $f(x) \neq 0$ which is given hence the limit exists.

3) According to arithmetic of continuous functions $\frac{(f(x))^2 +1}{f(x)}$ is valid because the denominator will never be $0$. Thus the $\frac{(f(x))^2 +1}{f(x)}$ is continuous at $x_0$.

Answer

we introduce g, such as $g(x)=f(x) + \frac{1}{f(x)}$, which is the sum of two functions .

We know that f is continuous at $x_0$ and non-null ( I guess), therefore, the inverse of this function is also continuous at that point.

The sum of two continuous functions at one point, is continuous, therefore g is continuous at $x_0$